Math Modeling Project December 3,2007

The Problem -The mountain reedbuck is a small deer common in the nature reserves of southern Africa. -The owner of a private game reserve in the Karoo has to cull a number of his stock of reedbuck to prevent overgrazing. -He is, however, not sure when and how many reedbuck must be removed, and is furthermore worried about the long-term effect of such a culling on his stock.

The Population Year Population 1988 260 1989 370 1990 500 1991 680 1992 950

Malthus Model N’[t]=kN[t] (Solved using Laplace Transforms) sn(s)-n(0)=kn(s) n(0)=alpha sn(s)-260=kn(s) alpha=260 sn(s)-kn(s)=260 n(s)(s-k)=260 n(s)=260/(s-k) n(s)=260*InverseLaplaceTransfrom[1/(s-k)] From Table on Page 149 N(t)=260e^(kt) or N(t)=Alpha*e^(kt)

Solving for the Constants N(t)=alpha*e^(kt) alpha=260 n(1)=370 370=260*e^(kt) (370/260)=e^(k) ln[370/260]=k .352821=k

The Model’s Results Year Population Model Predictions 1988 260 1989 370 1990 500 527 1991 680 749 1992 950 1066

Model Predictions

Culling Population using Model Introduce a minus E(t) into model. Table[NDSolve[{n'[t]=0.352821*n[t]-e,n[0]=a},n[t],{t,0,25}]]

E(t)=91 E(t)=92

Introducing E(t) as a Linear Function of Time. Model Minus E(t)*t E(t)=91*t E(t)=32*t

Conclusions about the Model The model first of all seems to overestimate the population. The model doesn’t seem to have a clear number of reedbucks to cull each year(91 grows the population exponentially while 92 kills the population off around 18 years). Another method needs to be used to find a better estimation of population of culling of reedbuck.

Logistic Model N’[t]=k*n[t]-s*n[t]^2 after a long derivation and simplifications the result given is as follows N[t]=[k/((k/alpha)-s)*e^(-kt)+s)]

Solving for the Constants N(1)=[k/(((k/260)-s)e^(-kt)+s))] 370*((k/260)-s)e^(-k)+s)=k (370k/260)e^(-k)-370se^(-k)+370s=k s(370-370e^(-k))=k-(370k/260)e^(-k) Equation 1 s=((k-(370k/260)e^(-k))/(370-370e^(-k)) Plug N(2) into equation and plug Equation 2 in for s and solve for k using mathematica Find Root Option k=0.486579 s=0.000428

The Model’s Results Year Population Model Predictions 1988 260 1989 370 1990 500 1991 680 638 1992 950 767

Model’s Prediction

Culling Population using Model Introduced a minus E(t) into model Used following equation NDSolve[{n’[t]==(0.486579n[t]-0.000428n[t]^2)-e ,n[0]==a},n[t],{t,0,25}]

Results of Model No Culling E(t)=92

Culling of Population E(t)=97 E(t)=98

Introducing E(t) as a Linear Function of Time. Model Minus E(t)*t E(t)=98*t E(t)=1*t

Conclusions about Model The model underestimates population and eventually limits the population around 1250 without culling. The model kills the population in 18 year if the culling is 98 or more per year. Overall the model seems reasonable in the short-term but long-term effects seem to be more questionable.

Overall Conclusions Both Models seem to closely agree with the culling amount for each year the models limits range from 92-98 (as long as culling remains constant and not a function of time). The culling could be affected by many other factors as well and the models could have some error based on constants but the best solution and long-term answer seem to be a culling of around 95 reedbuck per year. By taking an average of both models. The logistic model with the culling constant seems to have the least long-term effect on the culling of the population.